Chapter 2
Double and triple integration
(Stewart (Ed. 7): Chapter 15, p998.)
Chapter Summary
Objective
Double integration over regular domains
Tools
Sketch the domain of integration, decide if the domain is type I or type II or both and with the
aid of th
Chapter 1
Partial dierentiation
(Stewart (Ed. 7): Chapter 14, p901.)
Chapter Summary
Objective
Sketch and identify surfaces in 2-D and 3-D.
Find partial derivatives and use these to show simple
results, such as xux +yuy = 0. Deduce related second
order fo
Thursday, 25th October, 2012
1. Let f (x, y) =
Solutions to 2A (1 and 3) class test
exp(y)
f
f
. Find
and
.
4
2x + y
x
y
5
Solution:
f
0 2 exp(y)
2 exp(y)
=
=
4 )2
x
(2x + y
(2x + y 4 )2
f
(2x + y 4 ) exp(y) 4y 3 exp(y)
(2x + y 4 4y 3 )
=
= exp(y)
.
y
(2x
Tuesday, 25th October, 2011
10.10 a.m. to 10.55 a.m.
University of Glasgow
Department of Mathematics
Mathematics 2A - Multivariable Calculus
Class Test
An electronic calculator may be used provided that it does not have
a facility for either textual stora
Chapter 2
Double and triple integration
Example 2.1 Evaluate
x2 + y 2 dxdy
R
where R is [1, 3] [2, 4].
Solution
The integral may either be evaluated as
3
x2 + y 2 dxdy =
R
4
x2 + y 2 dy
dx
1
2
3
1
x2 y + y 3
3
=
1
3
(4 2)x2 +
=
1
=
2 3 56
x + x
3
3
4
dx
2
Chapter 4
Line and surface integrals: Solutions
Example 4.1 Find the work done by the force F(x, y) = x2 i xyj in moving a particle along the curve
which runs from (1, 0) to (0, 1) along the unit circle and then from (0, 1) to (0, 0) along the y-axis (see
Differentiation
Q1
Pre-course Revision
Find the derivatives of the following functions
a) y = (2x + 3)6 (Ans: 12(2x + 3)5 )
b) y =
c) y =
3
sin(2x )
(Ans: 6cosec(2x )cot(2x ).
1 )
2 x +7
x + 7 (Ans:
x3
).
cos2 x
d) y = x3 tan( x ) (Ans: 3x2 tan( x ) +
e)
Chapter 4
Line and surface integrals
Chapter Summary
Objective
Parametric equations of curves
Tools
The position vector of a point on the curve is
(x, y, z) = f (t), t R. This is called a parametric description of the curve and t is called a parameter. Th
Chapter 3
Dierentiation of vectors
Chapter Summary
Objective
Tools
Know and use the denition of grad, div and curl and
understand the meaning of vector and scalar elds
Nabla is the dierential operator,
For a scalar eld f , grad f =
(
=
f =
(
x , y , z
f f
Ex Sheet
5
2A Multivariable Calculus 2014
Tutorial Exercises
.
Lecture 9
Key Points:
Given the change of variables
T1
u=
1
( x + y)
3
denition of a Jacobian
v=
1
( x 2y)
3
use change of variables to evaluate double integrals
express x and y in terms of
Thursday, 6th December, 2012
4.30 p.m. to 6 p.m.
EXAMINATION FOR THE DEGREES OF
M.A. AND B.Sc.
Mathematics 2A - Multivariable Calculus
An electronic calculator may be used provided that it does not have
a facility for either textual storage or display or
Monday, 12th December, 2011
9.30 a.m. to 11.00 a.m.
2A degree exam - Solutions
1. (i) Solution:
z
z u z v
z 2
z
=
+
=
y exp(xy 2 ) +
.
x
u x v x
u
v
z
z u z v
z
z
z
=
+
=
2yx exp(xy 2 ) +
0=
2yx exp(xy 2 ).
y
u y v y
u
v
u
So the PDE becomes
2x
z
z 2
y ex
Thursday, 12th December, 2013
1.00 p.m. to 2.30 p.m.
EXAMINATION FOR THE DEGREES OF
M.A. AND B.Sc.
Mathematics 2A - Multivariable Calculus
An electronic calculator may be used provided that it does not have
a facility for either textual storage or display
?day, ?th December, 2013
? p.m. to ? p.m.
EXAMINATION FOR THE DEGREES OF
M.A. AND B.Sc.
Mathematics 2A - Multivariable Calculus
An electronic calculator may be used provided that it does not have
a facility for either textual storage or display, or for gr
Wednesday, 10th December, 2014
9.30 a.m. to 11 a.m.
EXAMINATION FOR THE DEGREES OF
M.A. AND B.Sc.
Mathematics 2A - Multivariable Calculus
An electronic calculator may be used provided that it does not have
a facility for either textual storage or display,
Tuesday, 9th November, 2010
Solutions to 2A class test
f
f
and
.
x
y
1. Let f (x, y) = x tan(xy + 2x). Find
5
Solution:
f
xy + 2x
= tan(xy + 2x) +
x
cos2 (xy + 2x)
x2
f
=
y
cos2 (xy + 2x)
2. By making the change of variables
x
and v = x,
y2
u=
nd the gene
Chapter 1
Partial dierentiation
Example 1.1 Sketch the graph of f (x, y) = 1 2x x2 y 2 .
Solution
: Let z = f (x, y). Completing the square, we have
z 2 = 1 2x x2 y 2 = 2 (x + 1)2 y 2 ,
i.e. (x + 1)2 + y 2 + z 2 = 2. This is the sphere with centre (1, 0,
5
Ex Sheet
2A Multivariable Calculus 2014
.
Tutorial Exercises
Given the change of variables
T1
u=
1
( x + y)
3
v=
1
( x 2y)
3
express x and y in terms of u and v.
Solution
x = 2u + v, y = u v.
Evaluate the following using the given change of variables:
T